c) (xy-1).(xy+5)

= x²y²+5xy-xy-5

=x²y²+4xy-5

a) b) d) bạn có thể ghi rõ được ko

d. ( x² - 2xy + y² ) ( x-y )
= ( x- y )² ( x- y )
= ( x - y )³

b, (x^2+xy+y^2) (x-y)
= ( x+ y )² ( x- y )
= ( x² - y ² ) ( x +y )

sau bạn đăng tách ra cho mn cùng giúp nhé 

a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)

b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)

c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)

d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)

e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)

f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)

đề bài là rút gọn à

`a, -xy(x^2+xy-y^2)`

`= -x^3y - x^2y^2 + xy^3`.

`b, 5x^2y(2y^2-xy)`

`= 10x^2y^3 - 5x^3y^2`.

`c, (-2x^3 - 1/4y - 4y^2).8xy^2`.

`= -16x^4y^2 - 2xy^3 - 32xy^4`.

`d, (2x^3 - 3xy + 12x)(-1/6xy)`

`= -2/3x^4y + 1/2x^2y^2 - 2x^2y`.

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

a)\(9y^3-y\)

\(=y\left(9y^2-1\right)\)

\(=y\left(3y-1\right)\left(3y+1\right)\)

\(9y^3-y=y\left(9y^2-1\right)=y\left(3y+1\right)\left(3y-1\right)\)

\(8y^3-2y\left(1-2y\right)^2=2y\left[\left(2y\right)^2-\left(1-2y\right)^2\right]=2y\left(4y-1\right)\)

\(2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\)

\(2x^4-6x^3+6x^2-2x=2x\left(x^3-3x^2+3x-1\right)=2x\left(x-1\right)^3\)\(x^3-8x^2+8x=x\left(x^2-8x+8\right)\)

\(5x^4-15x^3y+15x^2y^2-5xy^3-5x=5x\left(x^3-3x^2y+3xy^2-y^3-1\right)=5x\left[\left(x-y\right)^3-1\right]=5x\left(x-y-1\right)\left(x^2-2xy+y^2+x-y+1\right)\)

a)\(\left(3x^{3y}-\frac{1}{2}x^2+\frac{1}{5}xy\right).6xy^3\)

\(=18x^{3y+1}y^3-3x^3y^3+\frac{6}{5}x^2y^4\)

\(=y^3.\left(18x^{3y+1}-3x^3+\frac{6}{5}x^2y\right)\)

b)\(\frac{2}{3}x^{2y}.\left(3xy-x^2+y\right)\)

\(=2x^{2y+1}y-\frac{2}{3}x^{2y+x}+\frac{2}{3}x^{2y}y\)

c)(xy-1)(xy+5)

=x²y²+5xy-xy-5

=x²y²+4xy-5

d)Mk ko hiểu sao hai lần mũ liền

.a,(3x^3y - 1/2x^2 + 1/5xy) . 6xy^3

= 18x^4y^4 - 3x^3y^3 + 6/5x^2y^4

b, 2/3x^2y . (3xy - x^2 + y)

= 2x^3y^2 - 2/3x^4y^2 + 2/3x^2y^2

c, (xy - 1) . (xy + 5)

= x^2y^2 + 5xy - xy - 5

=x^2y^2 + 4xy - 5

d, (x^2y^2 - 1/2xy + 2xy)x - 2y  

= x^3y^2 - 1/2x^2y + 2x^2y - 2y